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3 years ago

Can someone explain to me how to do this!!!

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

Well...it says the sum of the angles measure 360...so lets set them equal to 360

40 + x - 10 + 1/3x + x - 20 = 360...combine like terms
2 1/3x + 10 = 360
7/3x = 360 - 10
7/3x = 350
x = 350/(7/3)
x = 350 * 3/7
x = 1050/7
x = 150 <=====

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3 years ago

Aria and her children went into a bakery and will buy donuts and brownies. She must buy at least 6 donuts and brownies altogethe

laila [671]

Answer:

Inequality for donuts and brownies is;

d ≥ 1 ≤ b

Step-by-step explanation:

We are told that aria will buy donuts and brownies.

We are also told that, number of donuts is d while number of brownies is b.

We are also told that she must by at least a total of 6 donuts and brownies together. Which means that the sum of d and b must not be 6 or greater than 6.

So the inequality will be;

d + b ≥ 6

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3 years ago

An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as

Paul [167]

Answer:

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is $\frac{a}{\sqrt{3}}=R$ .

Upon substituting our given values, we will get:

$\frac{5x}{\sqrt{3}}=6r$

Let us solve for r.

$r=\frac{5x}{6\sqrt{3}}$

$\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}$

We know that area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}s^2$ , where s represents side length of triangle.

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2$

The area within circle and outside the triangle would be difference of area of circle and triangle as:

$A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}$

We can make a common denominator as:

$A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}$

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Therefore, our required expression would be $A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$ .

7 0

3 years ago